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# mt_sn - EE 278 Statistical Signal Processing Wednesday...

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EE 278 Wednesday, November 11, 2009 Statistical Signal Processing Handout #16 Midterm Examination Solutions 1. Inequalities. a. e S is a convex function so by Jensen’s inequality E( e S ) e E( s ) = e Since E( S ) = 1. b. S is a concave function so by Jensen’s inequality E( S ) E( S ) 1 2 = 1 Since E( S ) = 1. 2. Exponentials. a. To find E( S | S > 5), we need to find the pdf of S given the event S > 5. Let event A = { S > 5 } . f S | A ( s ) = f S ( s ) P ( S A ) if s A , and 0 , otherwise = e - s P ( S > 5) = e - s integraltext 5 e - s ds = e - s e - 5 = e - ( s - 5) if s > 5, and 0 , otherwise Let s prime = s - 5 so s = s prime + 5. E( S | A ) = integraldisplay 5 sf S | A ( s ) ds = integraldisplay 5 se - ( s - 5) ds = integraldisplay 0 ( s prime + 5) e - s prime ds prime = integraldisplay 0 s prime e - s prime ds prime + integraldisplay 0 5 e - s prime ds prime = 1 + 5 = 6 b. To find E( S | S + T ), we first should recognize that E( S + T | S + T ) = E( S | S + T ) + E( T | S + T ) S + T = 2 E( S | S + T )

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Page 2 of 4 EE 278, Fall 2009 The first line follows from linearity of expectation. The second line follows because S and T are independent and have the same distribution so E( S | S + T ) = E( T | S + T ). Therefore, E( S | S + T ) = S + T 2 3. Waiting at the Bus Stop.
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